Question

A spring of original length $x$ is extended to thrice its length. If the force constant of the spring is $k\text{N/m}$, its potential energy is

• A. $\frac{k{x}^2}{2}$
• B. $\frac{3k{x}^2}{2}$
• C. $2k{x}^2$
• D. $\frac{5k{x}^2}{2}$

Answer 1

The correct option is C $2k{x}^2$

Potential energy of spring $=\frac{1}{2}k{x}^{\prime 2}$
Since the spring is extended to thrice its length,
$x^{\prime }=2x$ (where $x$ is the original length of the spring)

Thus, Potential energy $=\frac{1}{2}k{x}^{\prime 2}$
$=\frac{1}{2}k\times (2x{)}^2$
$=2k{x}^2$